I need help with the following derivatives
y= (e^1/x) / x^2
f(t) = t ln(5t)
thanks for your help
For dy/dx it appears you've not differentiated $\displaystyle e^{1/x}$ correctly
It's easier if you sub $\displaystyle u = \frac{1}{x}$ and then use the chain rule.
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For the second one remember the chain rule
$\displaystyle \frac{d}{dx} \ln [f(x)] = \frac{f'(x)}{f(x)}$
$\displaystyle y = \frac{e^{1/x}}{x^2}$
For this we need to use the quotient rule which is: $\displaystyle y' = \frac{vu' - uv'}{v^2}$
In your case $\displaystyle u = e^{1/x}$ and $\displaystyle v = x^2$
I will differentiate the numerator as that's what seems to be the issue then you can try using my value of $\displaystyle u'$ in the quotient rule. FYI my final answer is $\displaystyle \frac{dy}{dx} = \frac{e^{1/x}(2x-1)}{x^4} = -\frac{e^{1/x}(1-2x)}{x^4}$
To find u' (du/dx) we need to work out $\displaystyle \frac{d}{dx} e^{1/x}$.
For simplicity and clarity's sake let $\displaystyle t = \frac{1}{x}$. From the chain rule we know that $\displaystyle \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx}$
Due to this substitution we can rewrite u as $\displaystyle u = e^t$.
Differentiating this function is easy: $\displaystyle \frac{du}{dt} = e^t$
Similarly 1/x is also easy to differentiate: $\displaystyle \frac{dt}{dx} = -\frac{1}{x^2}$
From the chain rule above: $\displaystyle u' = \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx} = e^t \cdot \frac{1}{x^2}$
Since $\displaystyle t = \frac{1}{x}$ we get $\displaystyle u'$ in terms of x:
Therefore $\displaystyle u' = -\frac{1}{x^2}\,e^{1/x}$